注意head == None的情况，没用到Sorted这个特性，应该有占空间更少的解法。

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    # @param head, a ListNode
    # @return a ListNode
    def deleteDuplicates(self, head):
        if head == None:
            return None
        num = []
        cur = head
        while cur != None:
            if cur.val not in num:
                num.append(cur.val)
            cur = cur.next
        new_head = ListNode(num[0])
        cur = new_head
        for i in num[1:]:
            cur.next = ListNode(i)
            cur = cur.next
        return new_head